Problem: Simplify the following expression: $y = \dfrac{-7x^2+16x+15}{x - 3}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-7)}{(15)} &=& -105 \\ {a} + {b} &=& &=& {16} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-105$ and add them together. Remember, since $-105$ is negative, one of the factors must be negative. The factors that add up to ${16}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-5}$ and ${b}$ is ${21}$ $ \begin{eqnarray} {ab} &=& ({-5})({21}) &=& -105 \\ {a} + {b} &=& {-5} + {21} &=& 16 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-7}x^2 {-5}x) + ({21}x +{15}) $ Factor out the common factors: $ x(-7x - 5) - 3(-7x - 5)$ Now factor out $(-7x - 5)$ $ (-7x - 5)(x - 3)$ The original expression can therefore be written: $ \dfrac{(-7x - 5)(x - 3)}{x - 3}$ We are dividing by $x - 3$ , so $x - 3 \neq 0$ Therefore, $x \neq 3$ This leaves us with $-7x - 5; x \neq 3$.